Suppose we have a list of numbers called nums, we have to check whether every number can be grouped using one of the following rules: 1. Contiguous pairs (a, a) 2. Contiguous triplets (a, a, a) 3. Contiguous triplets (a, a + 1, a + 2)
So, if the input is like nums = [7, 7, 3, 4, 5], then the output will be True, as We can group [7, 7] together and [3, 4, 5] together.
To solve this, we will follow these steps −
n := size of nums
dp := a list of size n+1, first value is True, others are False
for i in range 2 to n, do
if i >= 2 and dp[i − 2] is not 0, then
if nums[i − 1] is same as nums[i − 2], then
dp[i] := True
if i >= 3 and dp[i − 3] is not 0, then
if (nums[i − 1], nums[i − 2], nums[i − 3]) are same or (nums[i − 1], nums[i − 2] + 1, nums[i − 3] + 2 are same), then
dp[i] := True
return dp[n]
Let us see the following implementation to get better understanding −
Example
class Solution: def solve(self, nums): n = len(nums) dp = [True] + [False] * n for i in range(2, n + 1): if i >= 2 and dp[i − 2]: if nums[i − 1] == nums[i − 2]: dp[i] = True if i >= 3 and dp[i − 3]: if (nums[i − 1] == nums[i − 2] == nums[i − 3]) or (nums[i − 1] == nums[i − 2] + 1 == nums[i − 3] + 2): dp[i] = True return dp[n] ob = Solution() nums = [8, 8, 4, 5, 6] print(ob.solve(nums))
Input
[8, 8, 4, 5, 6]
Output
True